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# Short Answer Type Questions(Part - 2) - Real Numbers Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions(Part - 2) - Real Numbers Class 10 Notes | EduRev

The document Short Answer Type Questions(Part - 2) - Real Numbers Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Ques 19: Prove that 5-2âˆš3 is an irrational number.
Sol: Let 5-2âˆš3 is a rational number
âˆ´ 5-2âˆš3= p/q  where p and q are co-prime integers and q â‰  0.

Since, p and q are integers.
âˆ´  p/2q is a rational number
i.e.,   is a rational number.
â‡’âˆš3 is a rational number.
But this contradicts the fact that âˆš3 is an irrational number.
So, our assumption that (5-2âˆš3) is a rational number is not correct.
âˆ´ (5-2âˆš3) is an irrational number.

Ques 20: Prove that (5+3âˆš2) is an irrational number.
Sol: Let (5+3âˆš2) is a rational number.
âˆ´ (5+3âˆš2) =  a/b [where â€˜aâ€™ and â€˜bâ€™ are co-prime integers and b â‰  0]

â‡’
â‡’
â‡’

â€˜aâ€™ and â€˜bâ€™ are integers,
âˆ´   is a rational number.
â‡’ âˆš2 is a rational number.
But this contradicts the fact that âˆš2 is an irrational number.
âˆ´ Our assumption that (5+3âˆš2) is a rational is incorrect.
â‡’ (5+3âˆš2) is an irrational number.

Ques 21: Use Euclidâ€™s Division Lemma to show that the square of any positive integer is either of the form â€˜3mâ€™ or â€˜3m + 1â€™ for some integer â€˜mâ€™.
Sol: Let x be a positive integer
âˆ´ x can be of the form 3p, (3p + 1) or (3p + 2)
When x = 3p,  we have
x2 = (3p)2
â‡’ x2 = 9p2
â‡’ x2  = 3 (3p2)
â‡’ x2 = 3m [Here 3p2 = m]
When x  = (3p + 1),  we have
x=  (3p + 1)2
â‡’ x2 = 9p2 + 6p + 1
â‡’ x2 = 3p (3p + 2) + 1
= 3m + 1,
Where m =p (3p +2)
When x   = 3p +2 ,  we have
x= (3p + 2)2
= 9p2 + 12p + 4
= 9p2 + 12p + 3 + 1
= 3 (3p2 + 4p + 1) + 1
= 3m + 1,  where
m  = 3p2 + 4p + 1
Thus, x2 is of the form 3m or 3m + 1.

Ques 22: Show that 2+âˆš3 is an irrational number.
Sol:
Let  2+âˆš3 is a rational number.

such that p and q are co-prime integers and q â‰  0.

Since, p and q are integers.
âˆ´   is rational
â‡’ âˆš3 is rational.
But this contradicts the fact that âˆš3  is an irrational.
âˆ´ Our assumption that (2+âˆš3) is a rational is incorrect.
Thus, (2+âˆš3) is an irrational.

Ques 23: Show that one and only one of n, n + 2 and n + 4 is divisible by 3.
Sol: Let us divide n by 3.
Let us get â€˜qâ€™ as quotient and â€˜râ€™ remainder.
âˆ´ n  = 3 Ã— q + r,where 0 â‰¤ r < 3
i.e., r  =  0, 1, 2
when r = 0,  then  n = 3q   ... (1)
when r = 1,  then  n = 3q + 1   ... (2)
when r = 2,  then  n = 3q + 2    ... (3)
From (1), n is divisible by 3
From (2), n = 3q + 1
Adding 2 to both sides, we have
n + 2 = (3q + 1) + 2
â‡’ n + 2 = 3q + 3
â‡’ n + 2 = 3 (q + 1)
3 (q + 1) is divisible by 3,
âˆ´  n + 2 is divisible by 3.
From (3),
n = 3q + 2
Adding 4 to both sides,
(n + 4) = (3q + 2) + 4
â‡’ n + 4 = 3q + 6 = 3 (q + 2)
3 (q + 2) is divisible by 3.
âˆ´  n + 4 is divisible by 3.
At one time, r has only one value out of 0, 1, 2. Hence, only one of n, n + 2, n + 4 is divisible by 3.
OR
Let q be an integer such that 3q, (3q + 1) or (3q + 2) is a positive integer. Let us consider the following cases :
Case-I: When n = 3q
3q Ã· 3, gives 0 as remainder
âˆ´  n = 3q is divisible by 3.
Next n = 3q â‡’ n + 2 = 3q + 2
(3q + 2) Ã· 3, gives 2 as remainder.
â‡’ n + 2 = (3q + 2) is not divisible by 3.
Again, n = 3q â‡’  n + 4 = 3q + 4 = (3q + 3) + 1
âˆ´ [(3q + 3) + 1] Ã· 3, gives 1 as remainder.
â‡’ n + 4 = (3q + 4) is not divisible by 3.
Thus, n is divisible by 3, but (n + 2) and (n + 4) are not divisible by 3.

Case-II: When n = 3q + 1
Since (3q + 1) Ã· 3, gives remainder as 1.
âˆ´ n = (3q + 1) is not divisible by 3.
Next, n + 2 = (3q + 1) + 2 = (3q + 3) + 0
[(3q + 3) + 0] Ã· 3, gives remainder as 0
â‡’ n + 2 = (3q + 1) + 2 is divisible by 3.
Again, n = (3q + 1) â‡’  n + 4 = (3q + 1) + 4 = (3q + 3) + 2
[(3q + 3) + 2] Ã· 3, gives remainder as 2
â‡’ n + 4 = (3q + 1) + 4 is not divisible by 3.
Thus, (n + 2) is divisible by 3, but n and (n + 4) are not divisible by 3.

Case-III: When n = 3q + 2
Since (3q + 2) Ã· 3, gives remainder as 2
âˆ´ n = 3q + 2 is not divisible by 3
Next n + 2 = (3q + 2) + 2 = (3q + 3) + 1
[(3q + 3) + 1] Ã· 3, gives remainder as 1.
â‡’ n + 2 = (3q + 2) + 2 is not divisible by 3
Again, n = 3q + 2   â‡’ n + 4 = (3q + 2) + 4 = (3q + 6) + 0
[(3q + 6) + 0] Ã· 3, gives remainder as 0
â‡’ n + 4 = (3q + 2) + 4 is divisible by 3
Thus, (n + 4) is divisible by 3, but n and (n + 2) are not divisible by 3.

Ques 24: Show that (2+âˆš5 )is an irrational number.
Sol:
Let (2+âˆš5)  is a rational number.
âˆ´ (2+âˆš5) = p/q  , such that p and q are co-prime integers and q â‰  0

p and q are integers.
âˆ´   is a rational.
â‡’ âˆš5 is a rational.
But, this contradicts the fact that âˆš5 is an irrational.
âˆ´ Our supposition that (2+âˆš5) is rational is incorrect.
Thus,(2+âˆš5)is an irrational.

Ques 25: Using Euclidâ€™s division algorithm, find the HCF of 56, 96 and 404.
Sol: Using the Euclidâ€™s division algorithm for 56 and 96
We have 96 = 56 Ã— 1 + 40  and
56 = 40 Ã— 1 + 16
Similarly 40 = 16 Ã— 2 + 8
and 16 = 8 Ã— 2 + 0
Remainder is zero
âˆ´ HCF of 56 and 40 is 8.
Now, 404 = 8 Ã— 50 + 4
and 8 = 4 Ã— 2 + 0
âˆµ Remainder is zero
âˆ´ HCF of 404 and 8, is 4
Thus, the HCF of 56, 96 and 404 is 4.

Ques 26: Prove that 3-âˆš5 is an irrational number.
Sol: Let (3-âˆš5) is a rational number.
âˆ´ 3-âˆš5 = p/q , such that p and q are co-prime integers and q â‰  0.

â‡’
â‡’

Since, p and q are integers,
âˆ´  is a rational number.
â‡’ âˆš5 is a rational number.
But this contradicts the fact that âˆš5 is an irrational number.
âˆ´ Our assumption that (3-âˆš5)  is a rational numberâ€™ is incorrect.
â‡’ (3-âˆš5)  is an irrational number.

Ques 27: Prove that (5+ âˆš2) is irrational.
Sol: âˆš2 = a/b where â€˜aâ€™ and â€˜bâ€™ are co-prime integers and b â‰  0

â‡’
â‡’

since â€˜aâ€™ and â€˜bâ€™ are integers,
âˆ´  is a rational.
â‡’ âˆš2 is a rational.
But this contradicts the fact that âˆš2 is an irrational.
âˆ´ Our assumption that (5+ âˆš2) is a rational number is incorrect.
Thus, (5+ âˆš2) is an irrational number.

Ques 28: Prove that 2âˆš3-7 is an irrational.
Sol: Let  2âˆš3-7 is rational.

â‡’
âˆµp and q are integers.
âˆ´   is rational.
â‡’ âˆš3 is rational.
But we know that âˆš3 is irrational,
âˆ´ Our assumption that (2âˆš3-7) is rational is wrong.
Hence 2âˆš3-7 is irrational.

Ques 29: If â€˜nâ€™ in an integer, then show that n2 â€“ 1 is divisible by 8.
Sol:
Le q be an integer, then
4q + 1 or 4q + 3
Now,
Case-I: When
n = 4q + 1
âˆ´ n= (4q + 1)2 = 16q2 + 8q + 1
â‡’ n2 â€“ 1 = 16q2 + 8q + 1 â€“ 1
= 16q2 + 8q
= 8q (2q + 1), which is divisible by 8.
â‡’ n2 â€“ 1 is divisible by 8.

Case-II: When
n = 4q + 3
âˆ´ n2  =  (4q + 3)2 = 16q2 + 24q + 9
â‡’ n2 â€“ 1=  16q2 + 24q + 9 â€“ 1 = 16q2 + 24q + 8
= 8(2q2 + 3q + 1) which is divisible by 8.
â‡’ n2 â€“ 1 is divisible by 8.
Thus, (n2 â€“ 1) is divisible by 8.

Ques 30: Prove that if x and y are both odd positive integers, then x2 + y2 is even but not divisible by 4.
Sol: Let â€˜qâ€™ be an integer, then any odd positive integer is of the form 2q + 1
Again let for some integers m and n are such that
x = 2m + 1 and y = 2n + 1
âˆ´ x2 + y2 = (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 +4n + 1
= 4(m2 + n2) + 4
(m + n) + 2
= 4[(m2 + n2) + (m + n)] + 2
= 4q + 2
[Where q = (m2 + n2) + (m + n)]
â‡’  x2 + y= 2[2q + 1], which an even number
Thus, 4q + 2 is an even number which is not divisible by 4, i.e. it leaves remainder 2.
Hence, x2 + y2 is even but not divisible by 4.

Ques 31: Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.
Sol: Let â€˜mâ€™ be an integer such that 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5 can be any positive integer.
âˆ´ An odd positive integer can be of the form 6m + 1, 6m + 3  or 6m + 5
Now, we have
(6m + 1) =  36m2 + 12m + 1
= 6(6m2 + 2m) + 1
= 6q + 1, q is an integer
(6m + 3)2  =  36m2 + 36m + 9
= 6(6m2 + 6m + 1) + 3
= 6q + 3, q is an integer
(6m + 5)= 36m2 + 60m + 25
= 6(6m2 + 10m + 4) + 1
= 6q + 1,  q is an integer
Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Ques 32: Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
Sol:
Let q be an integer such that 5q, (5q + 1) or (5q + 2) is a positive integer. Let us consider the following cases:
Case-I: When n = 5q
Since 5q Ã· 5, gives remainder as 0 â‡’ n is divisible by 5
Next, n = 5q â‡’ n + 4 = 5q + 4
â‡’ (n + 4) is not divisible by 5
and (5q + 4) Ã· 5, gives remainder as 4
Again,  n = 5q â‡’ n + 8 = 5q + 8 = (5q + 5) + 3
â‡’ (n + 8) is not divisible by 5
and [(5q + 5) + 3] Ã· 5, gives remainder as 3
Similarly,
n = 5q â‡’ n + 12 = 5q + 12 = (5q + 10) + 2
â‡’ (n + 12) is not divisible by 5
and [(5q + 10) + 2] Ã· 5, gives remainder as 2
n = 5q â‡’ n + 16 = 5q + 16 = (5q + 15) + 1
â‡’ (n + 16) is not divisible by 5
and [(5q + 15) + 1] Ã· 5, gives remainder as 1
Thus, n is divisible by 5, but (n + 4), (n + 8), (n + 12) and (n + 16) are not divisible by 5.

Case-II: When n = (5q + 1)
Here, (5q + 1) Ã· 5, gives remainder as 1 â‡’ n is not divisible by 5
Next  n = (5q + 1)
â‡’ n + 4 = (5q + 1) + 4 = (5q + 5) + 0
â‡’ (n + 4) is divisible by 5
and [(5q + 5) + 0] Ã· 5, gives remainder as 0
Similarly,
n = (5q + 1) â‡’ n + 8 = (5q + 1) + 8
= (5q + 5) + 4
â‡’ (n + 8) is not divisible by 5
and [(5q + 5) + 4] Ã· 5 gives remainder as 4
n = (5q + 1) â‡’ n + 12  = (5q + 1) + 12
= (5q + 10) + 3
â‡’ (n + 12) is not divisible by 5
and [(5q + 10) + 3] Ã· 5, gives remainder as 3
n = 5q + 1 â‡’ n + 16  = 5q + 1 + 16
= (5q + 15) + 2
â‡’ (n + 16) is not divisible by 5
and [(5q + 15) + 2] Ã· 5, gives remainder as 2
Thus, (n + 4) is divisible by 5, but n, (n + 8), (n + 12) and (n + 16) are not divisible by 5.

Case-III: When n = (5q + 2)
Here, (5q + 2) Ã· 5, gives remainder as 2
â‡’ n is not divisible by 5
Next n = 5q + 2 â‡’ n + 4 = 5q + 2 + 4
â‡’ 5q + 6 = (5q + 5) + 1
â‡’ (n + 4) is not divisible by 5
and [(5q + 5) + 1] Ã· 5, gives remainder as 1
Similarly,
n = 5q + 2 â‡’ n + 8 = 5q + 2 + 8 = 5q + 10
â‡’ (n + 8) is divisible by 5
and (5q + 10) Ã· 5, gives remainder as 0
n = 5q + 2 â‡’ n + 12 = 5q + 2 + 12
= (5q + 10) + 4
â‡’ (n + 12) is not divisible by 5
and [(5q + 10) + 4] Ã· 5, gives remainder as 4
n = 5q + 2 â‡’ n + 16 = 5q + 2 + 16
= (5q + 15) + 3
â‡’ (n + 16) is not divisible by 5
and [(5q + 15) + 3] Ã· 5, gives remainder 3
Thus, (n + 8) is divisible by 5, but n, (n + 4) (n + 12) and (n + 16) are not divisible by 5.

Case-IV: When n = (5q + 3)
Here, (5q + 3) Ã· 5, gives remainder as 3 â‡’ n is not divisible by 5.
Next n = 5q + 3 â‡’ n + 4 = (5q + 3) + 4
= (5q + 5) + 2
â‡’ (n + 4) is not divisible by 5
and [(5q + 5) + 2] Ã· 5, gives remainder as 2
Similarly,
n = 5q + 3 â‡’ n + 8 = 5q + 3 + 8
= (5q + 10) + 1
â‡’ (n + 8) is not divisible by 5
and [(5q + 10) +1] Ã· 5, gives remainder as 1
n = 5q + 3 â‡’ n + 12 = 5q + 3 + 12
= (5q + 15) + 0
â‡’ (n + 12) is divisible by 5
and [(5q + 15) + 0] Ã· 5, gives remainder as 0
n = 5q + 3 â‡’ n + 16 = 5q + 3 + 16
= (5q + 15) + 4
â‡’ (n + 16) is not divisible by 5
and [(5q + 15) + 4] Ã· 5, gives remainder as 4
Thus, (n + 16) is divisible by 5 but n, (n + 4) , (n + 8) and (n + 12) are not divisible by 5.

Case-V: When n = (5q + 4)
Here, (5q + 4) Ã· 5, gives remainder as 4 â‡’ n is not divisible by 5
Next, n = (5q + 4) â‡’ n + 4 = 5q + 4 + 4
= (5q + 5) + 3
â‡’ (n + 4) is not divisible by 5
and [(5q + 5) + 3] Ã· 5, gives remainder as 3
Similarly,
n = 5q + 4 â‡’ n + 8 = 5q + 4 + 8
= (5q + 10) + 2
â‡’ (n + 8) is not divisible by 5
and [(5q + 10) + 2] Ã· 5, gives remainder as 2
n = 5q + 4 â‡’ n + 12 = 5q + 4 + 12
= (5q + 15) + 1
â‡’ (n + 12) is not divisible by 5
and [(5q + 15) + 1] Ã· 5, gives remainder as 1
n = 5q + 4 â‡’ n + 16 = 5q + 4 + 16
= (5q + 20) + 0
â‡’ (n + 16) is divisible by 5
and [(5q + 20) + 0] Ã· 5, gives remainder 0
Thus, (n + 16) is divisible by 5, but n, n + 4, n + 8 and n + 12 are not divisible by 5.

Ques 33: Show that there is no positive integer â€˜pâ€™ for which is rational.
Sol:
If possible let there be a positive integer p for which  = a/b is equal to a rational i.e.  where a and b are positive integers.

Now

Also,

Since a, b are integers

are rationals

â‡’ (p + 1) and (p â€“ 1) are perfect squares of positive integers, which is not possible (because any two perfect squares differ at least by 3). Hence, there is no positive integer p for which  is rational.

Ques 34: Prove that is an irrational, where p and q are primes.
Sol: Let be rational
Let it be equal to â€˜râ€™
i.e.
Squaring both sides, we have

â‡’
â‡’   ...(i)
Since, p, q are both rationals
Also, r2 is rational (âˆµ r is rational)
âˆ´ RHS of (i) is a rational number
â‡’ LHS of (i) should be rational i.e.âˆšq  should be rational.
But âˆšq is irrational (âˆµ p is prime).
âˆ´ We have arrived at a contradiction.
Thus, our supposition is wrong.
Hence, âˆšp+âˆšq  is irrational.

Ques 35: Prove that âˆš2+âˆš3 is irrational.
Sol:
Let us suppose that âˆš2+âˆš3 is rational.
Also, let âˆš2+âˆš3 = a, where â€˜aâ€™ is rational
âˆ´ âˆš2 =  (a-âˆš3)
Squaring both sides, we have

â‡’
â‡’
â‡’
i.e.âˆš3 = a rational number

But it is a contradiction (âˆš3 is irrational)
Hence, âˆš2+âˆš3 is irrational.

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